# Probability and chess

1. ### fk378

367
1. The problem statement, all variables and given/known data
If 8 rooks (castles) are randomly places on a chessboard, compute the probability that none of the rooks can capture any of the others. That is, compute the probability that no row or file contains more than one rook.

3. The attempt at a solution
I just started it by knowing there are 64 squares on a chessboard. If there are 8 rooks, then that leaves 56 empty blocks.

Where to go from here?

2. ### g_edgar

607
How many different ways to put the rooks on the board without the no-capture restriction? How many ways with the restriction?

3. ### fk378

367
no-capture restriction: 8! ways
without the restriction: 56! ways

4. ### benorin

1,026
Not to give it away, but so you can check: P(no capture) $$\approx .911\times 10^{-5}$$.

5. ### fk378

367
I don't understand how that is? Can you explain?

6. ### Dick

25,893
Can you start by explaining why you think 56! is the number of ways to place them without restriction? That is certainly wrong.

7. ### Dunkle

56
First, place a rook in a random spot on the board. How many ways are there to do this? Next, figure out how many places are left on the board where the second rook can't take the first rook. Now, how many ways are there to place another rook on the board so that it can't take either of the first two. Continue this until you get to the last rook (there should only be one space left for that one). Finally, figure out how many possible ways there are to place the 8 rooks on the board with no restrictions.

8. ### Дьявол

365

So you need one rook in one row.

$$P(A)=\frac{m}{n}$$

8th row - the rook can move on 8! ways

Can you find the probability now?

Notice: you got 8 rooks so n=8 * ???

9. ### epenguin

2,332
The first sentence is not equivalent to the second. :tongue:

25,893
Why not?

11. ### epenguin

2,332
Because capture possibility depends on the colour of the rooks.